0.01 mole of iodoform `(CHI_(3))` reacts with Ag to produce a gas whose volume at NTP is
`2CHI_(3)+6Ag to6Agl(s)+C_(2)H_(2)(g)`

Options

A. 224 mL
B. 112 mL
C. 336 mL
D. None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \text{CHI}_3 + 6 \text{Ag} \rightarrow 6 \text{AgI} (s) + \text{C}_2\text{H}_2 (g) \]

### Step 2: Determine the moles of gas produced
From the balanced equation, we can see that:
- 2 moles of iodoform (CHI₃) produce 1 mole of acetylene (C₂H₂).

### Step 3: Calculate moles of C₂H₂ produced from 0.01 moles of CHI₃
Using the stoichiometry from the balanced equation:
- If 2 moles of CHI₃ yield 1 mole of C₂H₂, then:
\[ \text{Moles of C}_2\text{H}_2 = \frac{1}{2} \times \text{Moles of CHI}_3 \]
\[ \text{Moles of C}_2\text{H}_2 = \frac{1}{2} \times 0.01 = 0.005 \text{ moles} \]

### Step 4: Calculate the volume of C₂H₂ at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of 0.005 moles of C₂H₂ is calculated as follows:
\[ \text{Volume} = \text{Moles} \times \text{Volume per mole} \]
\[ \text{Volume} = 0.005 \text{ moles} \times 22.4 \text{ L/mole} \]
\[ \text{Volume} = 0.112 \text{ L} \]

### Step 5: Convert the volume to milliliters
To convert liters to milliliters:
\[ 0.112 \text{ L} = 0.112 \times 1000 \text{ mL} = 112 \text{ mL} \]

### Final Answer
The volume of gas produced at NTP is **112 mL**.

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